Two parallel planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$ have equations \(\mathbf{r}\cd

Question

Two parallel planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$ have equations $\mathbf{r}\cdot \left( \begin{matrix} 1 \ -3 \ 2 \end{matrix} \right)=-5$ and $\mathbf{r}\cdot \left( \begin{matrix}-1 \3 \-2 \end{matrix} \right)=-10$ respectively. Find the perpendicular distance between ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$, the exact cartesian equations of parallel planes which are $\sqrt{28}$ units away from ${{p}_{1}}$.

Accepted Answer

(i) $\frac{15}{\sqrt{14}}$units (ii) $x-3y+2z=14\sqrt{2}-5$ and $x-3y+2z=-14\sqrt{2}-5$

Vectors - Distance from a Point to a Plane

Vectors - Distance from a Point to a Plane

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